Inner product spaces of smooth functions

Question

In the space $C^1([0,1])$ where each $f$ is an element of the space and $$||f||= \left(\int_0^1\left(|f|^2+|f'|^2\right)dx \right)^{1/2}$$ How can it be shown that $||\cdot||$ is a norm of the space?

Answer 1

It is clear that $\|\lambda f\| = |\lambda| \|f\|$. It is also clear by continuity that that if$ \|f\| = 0$, then $f = 0$.

It remains to show the triangle inequality.

If $g\in C[0,1]^n$, then it is straightforward to show that $\|g\|_2 = \sqrt{\sum_k \int |g_k(t)|^2 dt}$ is a norm. In particular, if $g,h \in C[0,1]^n$, then $\|g+h\|_2 \le \|g\|_2+ \|h\|_2$.

Define $L:C^1[0,1] \to C[0,1]^2$ by $L(f) = (f,f')$. Then we see that $\|L(f)\|_2 = \|f\|$, where $\|\cdot\|$ is the norm in the question. If $f,g \in C^1[0,1]$, then $\|f+g\| = \|L(f+g)\|_2 = \|Lf + Lg\|_2 \le \|Lf\|_2 + \|Lg\|_2 = \|f\| + \|g\|$, as required.